NumberTheory with SageMath
Following exercises are from Fundamentals of Number Theory written by Willam J. Leveque.
Chapter 1 p. 5 prime pi(x): the number of prime numbers that are less than or equal to x. (same as π(x) in textbook.)
sage: prime_pi(10) 4 sage: prime_pi(10^3) 168 sage: prime_pi(10^10) 455052511
Also, you can see π(x) lim = 1 x→∞ x/ log(x) with following code.
sage: fori in range(4, 13): ....: print"%.3f" % (prime_pi(10^i) * log(10^i) / 10^i) ....: 1.132 1.104 1.084 1.071 1.061 1.054 1.048 1.043 1.039 p. 7 divisors(n): the divisors of n. number of divisors(n): the number of divisors of n (= τ(n)).
sage: divisors(12) [1, 2, 3, 4, 6, 12] sage: number_of_divisors(12) 6
1 For example, you may construct a table of values of τ(n) for 1 ≤ n ≤ 5 as follows:
sage: fori in range(1, 6): ....: print"t(%d):%d" % (i, number_of_divisors(i)) ....: t (1) : 1 t (2) : 2 t (3) : 2 t (4) : 3 t (5) : 2 p. 19 Division theorem: If a is positive and b is any integer, there is exactly one pair of integers q and r such that the conditions b = aq + r (0 ≤ r < a) holds. In SageMath, you can get quotient and remainder of division by ‘a // b’ and ‘a % b’.
sage: -7 // 3 -3 sage: -7 % 3 2 sage: -7 == 3 * (-7 // 3) + (-7 % 3) True p. 21 factor(n): prime decomposition for integer n.
sage: factor(12) 2^2 * 3 sage: for p,e in factor(12): ....: print p,e ....: 2 2 3 1 p. 24 sigma(n, k): The sum of kth powers of the positive divisors of |n|. For example,
sigma(12, 3) = 13 + 23 + 33 + 43 + 63 + 123 = 2044.
sage: sigma(12, 2) 210 sage: sigma(12, 3) 2044
2 Note that we can rewrite sigma(n, k) as
m Y k k kei (1 + pi + ··· pi ) i=1
e1 em when n = p1 ··· pm .
sage: def my_sigma(n, k): ....: prod =1 ....: for p,e in factor(n): ....: psum=0 ....: fori in range(0, e + 1): ....: psum+=p^(k*i) ....: prod*=psum ....: return prod ....: sage: my_sigma(12, 3) == sigma(12, 3) True
3 Chapter 2 p. 32 gcd(n, m): Finds the greatest common divisor of n and m.
sage: gcd(15, 21) 3 p. 33 xgcd(n, m): Finds bezout’s coefficient and the GCD of n and m.
sage: g, a, b = xgcd(123, 54) sage: gcd(123, 54) == g True sage: a * 123 + b * 54 == g True
Note that we can find bezout’s coefficient with extended euclidean algorithm.
sage: def my_xgcd(n, m): ....: if n == 0: ....: return (m, 0, 1) ifm>=0 else (-m, 0, -1) ....: else: ....: g,x, y = my_xgcd(m %n, n) ....: return (g, y - (m // n) * x, x) ....: sage: g, a, b = my_xgcd(123, 54) sage: gcd(123, 54) == g True sage: a * 123 + b * 54 == g True
Also you can see how euclidean algorithm works with following code.
sage: def my_euclidean(n, m): ....: if abs(m) < abs(n): ....: m,n=n,m ....: r = m % n ....: while r != 0: ....: print"%d=%d*%d+%d" % (m, n, m // n, r) ....: m,n=n,r ....: r = m % n ....: print"%d=%d*%d" % (m, n, m // n) ....: sage: my_euclidean(741, 715) 741 = 715 * 1 + 26 715 = 26 * 27 + 13 26 = 13 * 2
Note that gcd(741, 715) = 13. The implementation of extended euclidean algorithm with least remainder(p. 35) is left for readers.
4 p. 44 lcm(n, m): Finds the least common multiplier for n, m.
sage: lcm(15, 21) 105
You can check gcd(n, m) × lcm(n, m) = n × m.
sage: gcd(15, 21) * lcm(15, 21) == 15 * 21 True
5 Chapter 3
p. 50 IntegerModRing(n): Declare residue class with modulo n.
sage: R = IntegerModRing(10) sage: R(11) 1 sage: R(13) 3 sage: R(3) + R(8) 1 sage: R(3) * R(8) 4 sage: R(3)^-1 7
You can implement your own modular inverse function
invmod(a, n): Find b such that a × b ≡ 1 (mod n) with extended euclidean algorithm. p. 53 eulder phi(n): Computes the Euler’s totient function φ(n).
sage: euler_phi(17) 16 sage: euler_phi(36) 12 sage: 3^euler_phi(10) % 10 == 1 True
Note that aφ(n) ≡ 1 (mod n) when (a, n) = 1.
p. 54 You can check Theorem 3.8 as following.
sage: def my_euler_phi(n): ....: res = n ....: for p,e in factor(n): ....: res*=(1-1/p) ....: return res ....: sage: my_euler_phi(24) == euler_phi(24) True
6 p. 60 crt(a, b, m, n): Find a solution x to the congruences x ≡ a (mod m) and x ≡ b (mod n). crt(rs, ms): Find a simultaneous solution x to the congruences x ≡ ri (mod mi).
sage: crt([2, 3], [3, 4]) 11 sage: crt(2, 3, 3, 4) 11 sage: crt([3, 5, 7], [4, 21, 25]) 1307
Note that 1307 is a solution of following system of linear modular equations:
x ≡ 3 (mod 4) x ≡ 5 (mod 21) x ≡ 7 (mod 25) p. 63 IntegerModRing(n)[]: Declare polynomial ring of residue class with modulo n.
sage: R.
Also, you can use roots() to find roots of the polynomial.
sage: R.
7 Chapter 4 p. 79 primitive root(q): Calculates a primitive roots of q where q is 2, 4, pn or 2pn for a positive integer n and a prime p.
sage: primitive_root(17) 3
3 is primitive root in Z/17Z and we have
30 = 1, 31 = 3, 32 = 9, 33 = 10, 34 = 13, 35 = 5, 36 = 15, 37 = 11, 38 = 16, 39 = 14, 310 = 8, 311 = 7, 312 = 4, 313 = 12, 314 = 2, 315 = 1, 316 = 1
Also, you can find discrete logarithm of x to the base g in Z/qZ as following.
sage: p = 17 sage: R = IntegerModRing(p) sage: g = primitive_root(p) sage: x = R(g^10) sage: x.log(g) 10 p. 90 nth root(n): Find n-th roots of object if it exists.
sage: R = IntegerModRing(17) sage: g = R(primitive_root(17)) sage: g.nth_root(3)^3 == g True sage: g.nth_root(5)^5 == g True
8 Chapter 5 p. 99 kronecker(a, n): Calculates the Legendre symbol or it’s generelization, kronecker symbol.
sage: kronecker(2, 17) 1 sage: kronecker(14, 17) -1 sage: quadratic_residues(23) [0, 1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18] sage: [x forx in range(23) if kronecker(x, 23) == 1] [1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18]
You can check multiplicative property of Legendre symbol by
sage: def my_kronecker(a, n): ....: prod =1 ....: for p,e in factor(a): ....: ife%2== 0: ....: continue ....: else: ....: prod *= kronecker(p, n) ....: return prod ....: sage: my_kronecker(48, 31) == kronecker(48, 31) True sage: my_kronecker(48, 37) == kronecker(48, 37) True
To check the mulitiplicative property of Jacobi symbol,
sage: kronecker(2, 3) * kronecker(2, 5) == kronecker(2, 3 * 5) True
9 Chapter 6
p. 127 moebius(n): Calculate Moebius function µ(n). µ(n) = 0 if n is not square free, and otherwise equals (−1)r, where n has r distinct prime factors. For simplicity, µ(0) = 0 and µ(1) = 1.
sage: moebius(12) 0 sage: moebius(-5) -1 sage: moebius(-35) 1 sage: moebius(-7) -1 sage: moebius(4) 0
p. 131 floor(x): Find the largest integer not suceeding x. (Same as [x]) ceil(x): Find the smallest integer not preceding x.
sage: ceil(5.5) 6 sage: ceil(5) 5 sage: floor(5.5) 5 sage: floor(5) 5
p. 154 zeta(s): The Riemann zeta function.
sage: zeta(1) Infinity sage: zeta(2) 1/6* pi ^2 sage: zeta(2.) 1.64493406684823 sage: zeta(RealField(200)(2)) 1.6449340668482264364724151666460251892189499012067984377356 sage: zeta(0) -1/2
10 Chapter 7 p. 187 three squares(n): Write the integer n as a sum of three integer squares if possible.
sage: three_squares(389) (1, 8, 18) sage: three_squares(946) (9, 9, 28) sage: three_squares(2986) (3, 24, 49) four squares(n): Write the integer n as a sum of four integer squares.
sage: four_squares(3) (0, 1, 1, 1) sage: four_squares(13) (0, 0, 2, 3) sage: four_squares(130) (0, 0, 3, 11) sage: four_squares(1101011011004) (90, 102, 1220, 1049290)
11 Chapter 8 p. 198 (Pell’s equation) √ Let pi denote the sequence of convergents to the regular continued fraction for n. This se- qi quence is unique. Then the pair (x, y) solving Pell’s equation and minimizing x satisfies x = pi and y = qi for some i. This pair is called the fundamental solution. Thus, the fundamental solution may be found by performing the continued fraction expansion and testing each succes- sive convergent until a solution to Pell’s equation is found. continued fraction(x): Find continued faction of given real number. You can put nterms to limit the number of terms. numerator(n): Find the numerator of the n-th partial convergent of continued fraction. denominator(n): Find the numerator of the n-th partial convergent of continued fraction.
sage: continued_fraction(13/27) [0; 2, 13] sage: 0 + 1/(2 + 1/13) 13/27 sage: continued_fraction(sqrt(2)) [1; 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ...] sage: continued_fraction(sqrt(21)) [4; 1, 1, 2, 1, 1, 8, 1, 1, 2, 1, 1, 8, 1, 1, 2, 1, 1, 8, 1, ...] sage: continued_fraction(pi) [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...] sage: continued_fraction(pi, nterms = 5) [3; 7, 15, 1, 292]
Also, you can solve Pell’s equation by following:
sage: def solve_pell(N, numTry = 100): ....: cf = continued_fraction(sqrt(N)) ....: fori in range(numTry): ....: denom = cf.denominator(i) ....: numer = cf.numerator(i) ....: if numer^2 - N * denom^2 == 1: ....: return numer, denom ....: return None, None ....: sage: solve_pell(21) (55 , 12) sage: 55^2 - 21 * 12^2 1
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